Class 12th

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New answer posted

4 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

E? = φ + K?
E? = φ + K?
E? - E? = K? - K?
Now V? /V? = 2
K? /K? = 4; K? = 4K?
Now from equation (2)
⇒ 4 - 2.5 = 4K? - K?
1.5 = 3K?
K? = 0.5eV
Now putting this
Value in equation (2)
2.5 = φ + 0.5eV
φ = 2eV

New answer posted

4 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

Let us assume the potential at A = V_A = 0.
Now at junction C, according to KCL
i? + i? = i?
1 A + i? = 2 A


i? = 2 A
Now analyse potential along ACDB
V_A + 1 + i? (2) - 2 = V_B
0 + 1 + 2 (1) - 2 = V_B
V_B = 3 - 2
V_B = 1amp.

New answer posted

4 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Figure of Merit = C = i/θ
= C = (6 * 10? ³)/2 = 3 * 10? ³ Am²

New answer posted

4 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

x = 1/√ (µ? ε? ) = speed ⇒ [x] = [L¹T? ¹]
y = E/B = speed ⇒ [y] = [L¹T? ¹]
z = l/ (RC) = l/τ ⇒ [z] = [L¹T? ¹]
So, x, y, z all have the same dimensions.

New answer posted

4 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

1.52

E = E 0 - 0.0591 4 l o g ? H + 4

E = 1.23 + 0.0591 * p H

E = 1.23 + 0.0591 * ( 5 )

E = 1.52

New answer posted

4 months ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

4 months ago

0 Follower 9 Views

R
Raj Pandey

Contributor-Level 9

C o N H 3 6 C l 3 + 3 A g N O 3 ? 3 A g C l

  Mole of   C o N H 3 6 C l 3 1 =   Mole of   A g N O 3 3

0.3 267.46 = 0.125 * V * 10 - 3 3

V = 26.92 m L

New answer posted

4 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

Till input voltage reaches 4 V. No zener is in breakdown region. So V? = V? then now hen V? changes between 4 V to 6 V one zener with 4 V will breakdown are P.D. across this zener will become constant and remaining potential will drop acro resistance in series with 4 V zener.


Now current in circuit increases Abruptly and source must have an internal resistance due to which. Some potential will get drop across the source also so correct graph between V? and t will be
We have to assume some resistance in series with source.

New answer posted

4 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Number of bond between sulphur and oxygen = 8

Number of bond between sulphur and sulphur = 8

New answer posted

4 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Before inserting slab
C_i = ε? A/d
After inserting dielectric slab
C_i = ε? lw/d
C_f = C? + C?
C_f = (Kε? A? /d) + (ε? A? /d)
C_f = 2C_i ⇒ (Kε? wx/d) + (ε? w (l-x)/d) = 2ε? lw/d
4x + l - x = 2l
x = l/3

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